# Properties of Dedekind Domains

Throughout this article, A denotes a Dedekind domain.

Proposition 1.

Every fractional ideal of A can be written as

$\mathfrak m_1^{k_1} \ldots \mathfrak m_n^{k_n}, \quad k_1, \ldots, k_n \in \mathbb Z - \{0\}$

where each $\mathfrak m_i$ is a maximal ideal. The expression is unique up to permutation of terms.

Note

In particular, the maximal ideals of A generate its Picard group.

Proof

Existence. First suppose $\mathfrak a \subseteq A$ is a non-zero ideal. If $\mathfrak a = (1)$ we are done. Otherwise it is contained in some maximal ideal $\mathfrak m$, necessarily invertible. Now $\mathfrak a \ne \mathfrak a\mathfrak m^{-1}$, because if equality holds multiplying both sides by $\mathfrak {ma}^{-1}$ gives $\mathfrak m= A$, so $\mathfrak a \subsetneq \mathfrak a\mathfrak m^{-1} \subseteq A$. We replace $\mathfrak a$ by $\mathfrak a \mathfrak m^{-1}$ and repeat the process; this cannot continue indefinitely or we would have

$\mathfrak a \subsetneq \mathfrak a \mathfrak m_1^{-1}\subsetneq \mathfrak a \mathfrak m_1^{-1} \mathfrak m_2^{-1} \subsetneq \ldots$

Hence $\mathfrak a$ is a finite product of maximal ideals. Now a general fractional ideal is of the form $\frac 1 x \mathfrak a$ for a non-zero ideal $\mathfrak a \subseteq A$ and $x\in A-\{0\}$. Write both $\mathfrak a$ and $xA$ as a product of maximal ideals and we are done.

Uniqueness. Suppose $\mathfrak m_1^{k_1} \ldots \mathfrak m_n^{k_n} = \mathfrak m_1^{j_1} \ldots \mathfrak m_n^{j_n}$, where $j_i, k_i \in \mathbb Z$ (some possibly zero). If $(k_1, \ldots, k_n) \ne (j_1, \ldots, j_n)$, after cancelling terms and reordering, we obtain a relation of the form:

$\mathfrak m_1^{k_1} \ldots \mathfrak m_s^{k_s} = \mathfrak m_{s+1}^{k_{s+1}} \ldots \mathfrak m_t^{k_t}, \quad k_i \in \mathbb Z_{>0}.$

This is an equality of ideals. We get a contradiction because the LHS is contained in $\mathfrak m_1$ but the RHS is not. ♦

Exercise A

Prove that an integral domain is a PID if and only if it is a Dedekind domain and a UFD. Hence we have the following Venn diagram.

Now we have the following.

Proposition 2.

Let M, N be non-zero fractional ideals of A. Write, as in proposition 1,

$M = \mathfrak m_1^{d_1} \mathfrak m_2^{d_2} \ldots \mathfrak m_k^{d_k}, \quad N = \mathfrak m_1^{e_1} \mathfrak m_2^{e_2} \ldots \mathfrak m_k^{e_k}$

where each $\mathfrak m_i \subset A$ is maximal and $d_i, e_i \in \mathbb Z$. Then

\begin{aligned} M+N &= \mathfrak m_1^{\min(d_1, e_1)} \ldots \mathfrak m_k^{\min(d_k, e_k)}\\ M\cap N &= \mathfrak m_1^{\max(d_1, e_1)} \ldots \mathfrak m_k^{\max(d_k, e_k)}\\ MN &= \mathfrak m_1^{d_1 + e_1} \ldots \mathfrak m_k^{d_k + e_k},\\ (M:N) &= \mathfrak m_1^{d_1 - e_1} \ldots \mathfrak m_k^{d_k - e_k}\end{aligned}

Proof

Exercise. [ Localize at each $\mathfrak m_i$; reduce to the case where A is a dvr. ]

# Constructing Dedekind Domains

How do we find Dedekind domains? For starters, a PID is a Dedekind domain so we have $\mathbb Z$ and $k[X]$ (for any field k), but these are not too interesting.

Here is a standard way to construct new Dedekind domains from an existing one A. Let K be its field of fractions and L be a finite extension of K; we let B be the integral closure of A in L.

Exercise B

Prove that L is the field of fractions of B.

Now B has the following properties:

• it is a normal domain by construction;
• dim B = 1 since B is an integral extension of A, and dim A = 1 (by main theorem here).

Hence, if we can show that B is noetherian, then it is a Dedekind domain. But this result, called the Krull-Akizuki theorem, is surprisingly hard to prove. In most cases of interest, B will turn out to be a finitely generated A-module; thus B is noetherian as an A-module and hence as a module over itself.

Examples

1. Let $K/\mathbb Q$ be a finite extension and $\mathcal O_K$ be the integral closure of $\mathbb Z$ in K. One can show that $\mathcal O_K$ is a discrete subgroup of $K$, when we topologize $K$ by identifying it with $\mathbb Q^n$. Hence $\mathcal O_K$ is a finitely generated $\mathbb Z$-module; in particular it is a noetherian ring and hence a Dedekind domain. A more general result of this nature will be proven later.

2. Let k be any field such that $\mathrm{char} k \ne 2$. The integral closure of $k[X]$ in $L = k(X)[Y]/(Y^2 - X^3 + X)$ is $A = k[X, Y]/(Y^2 - X^3 +X)$. [ Proof: exercise. ] Since A is clearly noetherian, it is a Dedekind domain.

Exercise C

Let $m\in \mathbb Z - \{\pm 1\}$ be a square-free integer and $K = \mathbb Q(\sqrt m)$, which is quadratic over $\mathbb Q$. Prove that

$\mathcal O_K = \begin{cases} \mathbb Z[\frac{1 + \sqrt{m}}2], \quad &\text{if } m\equiv 1 \pmod 4, \\ \mathbb Z[\sqrt{m}], \quad &\text{if } m\equiv 2,3 \pmod 4.\end{cases}$

# Valuation

Definition.

Let $K = \mathrm{Frac} A$. Each maximal $\mathfrak m \subset A$ gives a function

$\nu_{\mathfrak m} : K - \{0\} \longrightarrow \mathbb Z$,

which takes $x$ to the exponent of $\mathfrak m$ in the above factorization of $xA$. This is called the valuation of $x$ at $\mathfrak m$.

Exercise D

1. Prove the following properties of $\nu = \nu_{\mathfrak m}$: for any $x,y\in K - \{0\}$ we have:

$\nu(1) = 0, \quad \nu(xy) = \nu(x) + \nu(y), \quad \nu(x+y) \ge \min(\nu(x), \nu(y)) \text{ if } x+y\ne 0$.

Verify that if we set $\nu(0) = \infty$ these properties hold for all $x,y\in K$.

2. Define a distance function on K by $d(x, y) = e^{-\nu(x-y)}$. Prove that (Kd) is a metric space, where the metric satisfies the strong triangular inequality.

$x,y, z\in K \implies d(x, z) \le \max(d(x, y), d(y, z)).$

A metric which satisfies this inequality is called an ultrametric. Interpret and prove the following statement: in an ultrametric space, every triangle is isoceles.

## Valuation in Geometry

Suppose $A = \mathbb C[X, Y]/(Y^2 - X^3 + X)$, a Dedekind domain. Recall that points P on the curve $Y^2 = X^3 - X$ correspond bijectively to maximal ideals $\mathfrak m_P \subset A$. Then $\nu_{\mathfrak m_P}(f)$ is just the order of vanishing of at P.

For example, let $\mathfrak m = (X, Y) = \mathfrak m_P$ where $P = (0, 0)$; for convenience write $\nu := \nu_{\mathfrak m}$. Let us compute $\nu(f)$ for $f = X^2 - Y^4 \in A$:

$X^2 - Y^4 = X^2 - (X^3 - X)^2 = 2X^4 - X^6 = X^4 ( 2 - X^2).$

Now $\nu(2 - X^2) = 0$ since f does not vanish on P. From an earlier example, we have $\mathfrak m_P^2 = (X)$ so $\nu(X) = 2$ and $\nu(X^2 - Y^4) = 8$.

As a simple exercise, compute $\nu(Y)$.

# Composition Series

Recall in the last example here that for distinct maximal ideals $\mathfrak m_1, \ldots, \mathfrak m_n$ of any ring A and $k_1, \ldots, k_n \ge 1$ we have a ring isomorphism

$A/(\mathfrak m_1^{k_1} \ldots \mathfrak m_n^{k_n}) \cong A/\mathfrak m_1^{k_1} \times \ldots \times A/\mathfrak m_n^{k_n}.$

This gives the following.

Proposition 3.

If A is a Dedekind domain with finitely many maximal ideals $\mathfrak m_1, \ldots, \mathfrak m_n$, then it is a PID.

Proof

It suffices to prove that each $\mathfrak m_i$ is principal. Let us consider $\mathfrak m_1$. Since $\mathfrak m_i^2 \subsetneq \mathfrak m_i$ we can pick $y \in \mathfrak m_1 - \mathfrak m_1^2$. From the isomorphism

$A/(\mathfrak m_1 \ldots\mathfrak m_n) \cong A/\mathfrak m_1 \times \ldots \times A/\mathfrak m_n$,

we can find $x\in A$ such that

$x\equiv y \pmod {\mathfrak m_1}, \ x \equiv 1 \pmod {\mathfrak m_2}, \ \ldots, \ x \equiv 1 \pmod {\mathfrak m_n}.$

Then x is divisible by $\mathfrak m_1$ but not its square, and not divisible by $\mathfrak m_2, \ldots, \mathfrak m_n$. Thus $\mathfrak m_1 = (x)$. ♦

Proposition 4.

Let $\mathfrak a \subseteq A$ be a non-zero ideal; write

$\mathfrak a = \mathfrak m_1^{k_1} \ldots \mathfrak m_n^{k_n}.$

Then the composition factors of $A/\mathfrak a$ as an A-module comprise of $k_i$ copies of $A/\mathfrak m_i$ for $1 \le i \le n$. In particular

$l_A(A/\mathfrak a) = \sum_{i=1}^n k_i$.

Note

Recall that the composition factors of $A/\mathfrak a$ as an A-module are identical to those as a module over itself.

Proof

The above ring isomorphism

$A/(\mathfrak m_1 \ldots\mathfrak m_n) \cong A/\mathfrak m_1 \times \ldots \times A/\mathfrak m_n$

is in fact an isomorphism of A-algebras. Hence $l_A(A/\mathfrak a) = \sum_i l_A(A/\mathfrak m_i^{n_i})$. We obtain a series of submodules of $A/\mathfrak m^n$

$0 \subsetneq \mathfrak m^{n-1}/\mathfrak m^n \subsetneq \mathfrak m^{n-2}/\mathfrak m^n \subsetneq \ldots \subsetneq \mathfrak m/\mathfrak m^n \subsetneq A/\mathfrak m^n$

with successive quotients $\mathfrak m^i/\mathfrak m^{i+1}$. It suffices to show each of these is of dimension 1 over $A/\mathfrak m$. To prove that, we recall that $A_{\mathfrak m}$ is a dvr so $\mathfrak n := \mathfrak m A_{\mathfrak m}$ is a principal ideal. Since $\mathfrak m^i / \mathfrak m^{i+1}$ is already a module over $A_{\mathfrak m}$ we have:

$\dim_{A/\mathfrak m} \mathfrak m^i/\mathfrak m^{i+1} = \dim_{A_{\mathfrak m}/\mathfrak n} \mathfrak n^i / \mathfrak n^{i+1} = 1$

as desired. ♦

Note

Consider the special case where $A = k[V]$ is the coordinate ring of a variety V over an algebraically closed field k. If A is a Dedekind domain, then since $A/\mathfrak m \cong k$ for each maximal ideal $\mathfrak m \subset A$ we have

$l_A(A/\mathfrak a) = \dim_k A/\mathfrak a$.

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